Let $X_n$ be the total number of people in a population at time $n$ and let $Y_i \sim \text{Bin}(2, p)$ be iid random variables for $i = 1, 2,...$ and $p>1/2$.
Consider the Markov chain such that $X_n = X_{n-1} + Y_{n} - 1$ with state space $\{0, 1, 2, .... \} $and assume that $X_0 = 1$.
What is the probability $X_n$ eventually gets absorbed to state $0$?
My Attempt:
I was confused between whether this would be a Galton-Watson Branching process problem or a markov chain (like a random walk/gambler's ruin) or something else.
I tried $\mathbb{P}(\bigcup_{n=1}^\infty (X_n = 0)|X_0 = 1) = \sum_{x=0}^2 \mathbb{P}(\bigcup_{n=1}^\infty (X_n=0)|X_1 = x)\mathbb{P}(X_1=x|X_0=1).$
Another thought was to try finding the stationary distribution but having "infinite sized transition matrix" I couldn't see how it would work.
But I couldn't get any further than this.
Thanks
It’s not a branching process; a branching process has increments according to descendants for each existing individual, whereas here the increments $Y_n$ are independent of $X_{n-1}$.
It’s a Markov chain because the increments don’t depend on the state history. (That’s also true for a branching process, so a branching process is also a Markov chain.)
You can ignore the state transitions where the state remains unchanged and just look at the transitions $k\to k-1$ with probability $(1-p)^2$ and $k\to k+1$ with probability $p^2$. Then you have an asymmetric simple random walk on the non-negative integers with probability ratio $\left(\frac{1-p}p\right)^2$.
Since the walk is invariant under translations, the probability to reach $1$ from $2$ is the probability to reach $0$ from $1$. Denote this probability by $q$. From $1$ you can either go to $0$ with probability $\frac{(1-p)^2}{p^2+(1-p)^2}$, or you can go to $2$ with probability $\frac{p^2}{p^2+(1-p)^2}$ and then return to $1$ with probability $q$ to get another chance at getting to $0$. Thus
$$ q=\frac{(1-p)^2}{p^2+(1-p)^2}+\frac{p^2}{p^2+(1-p)^2}\cdot q^2\;, $$
which has the form
$$ q=(1-r)+rq^2\;, $$
with solution
$$ q=\frac{1-r}r=\left(\frac{1-p}p\right)^2\;. $$